A344941 Numbers that are the sum of five fourth powers in exactly six ways.

151300, 225890, 236194, 243235, 246674, 250834, 286114, 288579, 300835, 302130, 302210, 303235, 309059, 317795, 320195, 334819, 334899, 335443, 336210, 338914, 346835, 356899, 363379, 366995, 373234, 375619, 389875, 391154, 392259, 393314, 394354, 412339

Offset: 1

Author

David Consiglio, Jr., Jun 03 2021

Keywords

• nonn

Comments

Differs from A344940 at term 2 because 197779 = 1^4 + 5^4 + 6^4 + 16^4 + 19^4 = 1^4 + 7^4 + 11^4 + 12^4 + 20^4 = 1^4 + 10^4 + 12^4 + 17^4 + 17^4 = 2^4 + 4^4 + 5^4 + 7^4 + 21^4 = 3^4 + 5^4 + 6^4 + 6^4 + 21^4 = 4^4 + 7^4 + 9^4 + 13^4 + 20^4 = 11^4 + 13^4 + 14^4 + 15^4 + 16^4.

Examples

			151300 is a term because 151300 = 3^4 + 3^4 + 3^4 + 12^4 + 19^4  = 3^4 + 11^4 + 11^4 + 14^4 + 17^4  = 3^4 + 13^4 + 13^4 + 13^4 + 16^4  = 6^4 + 9^4 + 9^4 + 9^4 + 19^4  = 7^4 + 11^4 + 11^4 + 11^4 + 18^4  = 8^4 + 9^4 + 13^4 + 13^4 + 17^4.
		

Links

• David Consiglio, Jr., Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A344359, A344921, A344940, A344943, A345175, A345818.

Programs

Python

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 for x in range(1, 1000)]
for pos in cwr(power_terms, 5):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 6])
for x in range(len(rets)):
    print(rets[x])